integral from 0 to 2 of (x^2+1)^2

Lösung

\int_{0}^{2} (x^{2} + 1)^{2} d x

\frac{206}{15}

Dezimale

13.73333 \dots

Schritte zur Lösung

\int_{0}^{2} (x^{2} + 1)^{2} d x

Multipliziere aus

(x^{2} + 1)^{2} : x^{4} + 2 x^{2} + 1

= \int_{0}^{2} x^{4} + 2 x^{2} + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2} x^{4} d x + \int_{0}^{2} 2 x^{2} d x + \int_{0}^{2} 1 d x

\int_{0}^{2} x^{4} d x = \frac{32}{5}

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

\int_{0}^{2} 1 d x = 2

= \frac{32}{5} + \frac{16}{3} + 2

Vereinfache

\frac{32}{5} + \frac{16}{3} + 2 : \frac{206}{15}

= \frac{206}{15}

\int_{0}^{1} e^{- x} d x

\int_{0}^{4} x + 2 d x

\int_{2}^{4} (- x^{2} + 2 x + 8) d x

\int_{0}^{1} (x^{12} + 1 2^{x}) d x

\int_{0}^{1} x^{4} - x^{5} d x